I know that the first step is to find the factors of 6 and 2, then see which when multiplied by the other coefficients have them add up to equal zero, but none of the factors I tried came out to zero. Is there an easier way to go about doing this???

4 Answers
4

Your method will in general not find all real, but only all rational zeroes.
If the leading coefficient were 1 instead of 2, all rational zeroes would have to be divisors of 6 (i.e. $\{\pm1, \pm2, \pm3, \pm6\}$).
However with a leading coefficient of 2, one should also check halves of these values (i.e. also {$\pm\frac12, \pm\frac32\}$).
Plugging in $x=2$, you will find that it is in fact a root. By polynomial division you thus obtain a quadratic for the other roots, which you can solve (or you will happen to find the remaining roots also by trying the above candidates).

find a number x that when you substitute in the polynomial will make it zero: the factor theorem. E.g when i substitute x = 2 , it gives zero. therefore (x-2) is a factor of the polynomial, then you perform polynomial division. when you divide 2x^2+x^2-13x+6 by (x-2) we get 2x^2+5x-3. then factorise 2x^2+5x-3 we get (x+3)(2x-1). therefore zeros are X=2,-3 and 1/2